Clearly though $(\sum_{i=1}^{2n} a_i (-1)^i)^2 \geq 0$ so we can't simply move on from this step, I feel I need to back up a bit to continue, but i've got a bit lost, if anyone could help it'd be great!

EDIT: Condition for equality. For equality to occur, we must have equality in both the applications of Cauchy-Schwarz in $\color{Red}{(\ast)}$. This is possible if and only if $a_1 = a_3 = a_5 = \cdots = a_{2n-1}$ and $a_2 = a_4 = \cdots = a_{2n-2} = a_{2n}$; i.e., iff there exist $c_0$ and $c_1$ such that $a_i = c_{i\ \bmod 2}$ for all $i$.

Here is an expansion of my comment, since it gives a different solution than Srivatsan's.

Let $A=(a_i), B=(b_i),C=(c_i)$ with $b_i=1,c_i=(-1)^n$. Then $\|B\|^2=\|C\|^2=2n$, and $B\cdot C = 0$. Let $B'=B/\|B\|$ and $C'=C/\|C\|$, and let $V$ be the vector space spanned by $B$ and $C$. Then we have the following inequalities: